$ A = \left[\begin{array}{rr}-1 & 5 \\ 3 & -2 \\ 1 & -1\end{array}\right]$ $ E = \left[\begin{array}{rr}3 & -2 \\ 1 & 4\end{array}\right]$ What is $ A E$ ?
Solution: Because $ A$ has dimensions $(3\times2)$ and $ E$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ A E = \left[\begin{array}{rr}{-1} & {5} \\ {3} & {-2} \\ \color{gray}{1} & \color{gray}{-1}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{-2} \\ {1} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{1} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{1} & ? \\ {3}\cdot{3}+{-2}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{1} & {-1}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{4} \\ {3}\cdot{3}+{-2}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{1} & {-1}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{4} \\ {3}\cdot{3}+{-2}\cdot{1} & {3}\cdot\color{#DF0030}{-2}+{-2}\cdot\color{#DF0030}{4} \\ \color{gray}{1}\cdot{3}+\color{gray}{-1}\cdot{1} & \color{gray}{1}\cdot\color{#DF0030}{-2}+\color{gray}{-1}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}2 & 22 \\ 7 & -14 \\ 2 & -6\end{array}\right] $